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Schach [20]
3 years ago
14

Why are hypothesis valuable even if it is not supported by data.

Physics
2 answers:
KengaRu [80]3 years ago
8 0
A hypothesis proposes a mechanism of change that can be tested. If the results indicate the hypothesized mechanism is not the one in operation it narrows the field and we know more. A falsified hypothesis tested by a well designed experiment still provides real evidence about the world. 

<span>It was once thought that maggots spontaneously arose from carrion. This was tested and falsified providing evidence this was not how new generations of flies appeared. This still left the question of how flies generated new life to be explored. One falsified hypothesis does not block further research into the question.

</span><span>Even a failed hypothesis can increase scientific knowledge if done correctly. By knowing this hypothesis is not how something works, you save time for future scientists and can now use your effort in other investigations. That is the real value of hypothesis that are not supported by evidence.</span>
disa [49]3 years ago
5 0
A hypothesis is an educated guess. Your data might not match your hypothesis because like i said, a hypothesis is a guess. Your data is multiple experiments to prove whether your guess was right or wrong. The hypothesis is valuable because because it gives you a reason to prove something right or wrong. Not sure how helpful that was but it's what i can think of...
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3 years ago
A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −
romanna [79]

Answer:

The speed of electron is 1.5\times10^{7}\ m/s

Explanation:

Given that,

Electric field E=1.50\times10^{3}\ N/C

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The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy P.E=5.04\times10^{-17}\ J

Change in potential energy \Delta P.E=-9.60\times10^{-17}\ J as it goes x = 0.190 m to x = -0.210 m

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W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))

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Using change in kinetic energy,

\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2

\Delta K.E=\dfrac{-3mv^2}{8}

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\Delta K.E=W

\Delta K.E=\Delta U

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Put the value of m

v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}

v=1.21\times10^{7}\ m/s

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Using given potential energy

\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})

\Delta U=-4.56\times10^{-17}\ J

We need to calculate the speed of electron

Using change in energy

\Delta U=-W=-\Delta K.E

\Delta K.E=\Delta U

\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}

Put the value into the formula

v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}

v_{f}=1.5\times10^{7}\ m/s

Hence, The speed of electron is 1.5\times10^{7}\ m/s

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